(a) The electrical power demand of TATA Power Corporation from 2018 to 2024, measured in megawatts, is provided below. Using the data, forecast the demand for the year 2025 by fitting and plotting the linear trend line.

Year | Electrical Power Deman

Question

(a)

The electrical power demand of TATA Power Corporation from 2018 to 2024, measured in megawatts, is provided below. Using the data, forecast the demand for the year 2025 by fitting and plotting the linear trend line.

Year | Electrical Power Demand (Megawatts) 2018 | 94 2019 | 99 2020 | 100 2021 | 110 2022 | 125 2023 | 162 2024 | 142

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Year | Electrical Power Demand (Megawatts)
2018 | 94
2019 | 99
2020 | 100
2021 | 110
2022 | 125
2023 | 162
2024 | 142

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(b) In a study to evaluate the effectiveness of two different animal feeds, researchers recorded the weight gains (in kilograms) of animals after fed each type of feed.

Feed A | 45 | 48 | 47 | 50 | 46 | 49 | 47 | 48
Feed B | 51 | 53 | 52 | 55 | 50 | 52 | 54 | 33

(i) If the two samples are considered independent, can we conclude that Feed B is more effective than Feed A at the 0·025 level of significance ?

(ii) Also examine the case if the same set of eight animals received both feeds, and test at the 0·01 level of significance. (Table enclosed)
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(c) Techline Pvt. Ltd. is a consumer electrics manufacturing concern producing three models of Smart Devices : X, Y and Z. The company operates under the constraints of two key resources, viz. raw materials and labor hours. Each week, the availability of raw materials is restricted to 800 kg, while labor availability is limited to 600 hours. To maximize profit, the company needs to decide, how many units of each product to manufacture weekly without exceeding the available resources. The relevant details for each product are provided below :

| Product | Raw Material per unit (Kg) | Labor Hours per unit (Hours) | Profit contribution per unit (₹) |
|---------|---------------------------|------------------------------|----------------------------------|
| X | 6 | 5 | 80 |
| Y | 4 | 4 | 60 |
| Z | 2 | 3 | 40 |

(i) Formulate this scenario as a Linear Programming Problem (LPP), and solve it for optimal product mix.

(ii)

Based on the optimal solution, analyze the situation if the company wants to introduce a New product W, requiring 2 kg of raw materials and 2 hours of labor hours per unit. Its estimated profit contribution is ₹35/- per unit.

Should the company include this new product in its production mix ? 20 marks

Should the company include this new product in its production mix ?
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UPSC Answer Check · Accepted Answer

(a) Let x = year − 2021 and y = demand in MW. The data become: x = −3, −2, −1, 0, 1, 2, 3; y = 94, 99, 100, 110, 125, 162, 142. n = 7, Σx = 0, Σy = 832, Σxy = 295, Σx² = 28. By the least-squares method, the linear trend is y = a + bx, where b = Σxy/Σx² = 295/28 = 10.5357, a = Σy/n = 832/7 = 118.8571. So the fitted trend line is: y = 832/7 + (295/28)x = 118.8571 + 10.5357(year − 2021). For 2025, x = 4: y = 832/7 + (295/28)(4) = 832/7 + 295/7 = 1127/7 = 161. Forecast demand for 2025 = 161 MW. Plot: plot the points (2018,94), (2019,99), (2020,100), (2021,110), (2022,125), (2023,162), (2024,142). ```svg ``` The straight line passes through (2021,118.86) and (2025,161.00); at x = −3 it gives 87.25 MW and at x = 3 it gives 150.46 MW. This assumes the underlying trend remains linear. (b) (i) Let H₀: μB − μA ≤ 0 and H₁: μB − μA > 0. Use the independent two-sample t-test, assuming equal variances. Feed A: n₁ = 8, ΣA = 380, mean A = 47.5, Σ(A − mean)² = 18, sA² = 18/7. Feed B: n₂ = 8, ΣB = 400, mean B = 50.0, Σ(B − mean)² = 348, sB² = 348/7. Pooled variance: s_p² = [(n₁ − 1)sA² + (n₂ − 1)sB²]/(n₁ + n₂ − 2) = (18 + 348)/14 = 366/14 = 183/7. Standard error: SE = √[s_p²(1/n₁ + 1/n₂)] = √[(183/7)(1/8 + 1/8)] = √(183/28) = 2.5565. t = (mean B − mean A)/SE = (50.0 − 47.5)/2.5565 = 2.5/2.5565 = 0.978. At α = 0.025, one-tailed, df = 14, t-critical = 2.145. Since 0.978 < 2.145, we fail to reject H₀. No, we cannot conclude that Feed B is more effective than Feed A at the 0.025 level. (ii) If the same eight animals received both feeds, use the paired t-test. Differences d = B − A: 6, 5, 5, 5, 4, 3, 7, −15. n = 8, Σd = 20, mean d̄ = 20/8 = 2.5. Σ(d − d̄)² = 360, so s_d² = 360/7 = 51.4286. SE = s_d/√n = √(360/7)/√8 = √(360/56) = √(45/7) = 2.5355. t = d̄/SE = 2.5/2.5355 = 0.986. At α = 0.01, one-tailed, df = 7, t-critical = 2.998. If two-tailed, t-critical = 3.499. Since 0.986 < 2.998, we fail to reject H₀. Even for paired data, we cannot conclude at the 0.01 level that Feed B is more effective. (c) (i) Let x, y, z be weekly units of X, Y, Z. Maximise P = 80x + 60y + 40z subject to: 6x + 4y + 2z ≤ 800 (raw material) 5x + 4y + 3z ≤ 600 (labour hours) x, y, z ≥ 0. The raw-material constraint is redundant, because 6x + 4y + 2z ≤ 6x + 4.8y + 3.6z = 1.2(5x + 4y + 3z) ≤ 1.2(600) = 720 < 800. Thus only labour is effectively binding. Profit per labour hour: X = 80/5 = 16, Y = 60/4 = 15, Z = 40/3 = 13.33. So labour is best used on X. If only X is produced: 5x ≤ 600 ⇒ x = 120. Raw material used = 6(120) = 720 ≤ 800. Feasible. Profit = 80(120) = ₹9,600. Optimal mix: x = 120, y = 0, z = 0. Maximum weekly profit = ₹9,600. Raw material slack = 800 − 720 = 80 kg; labour slack = 0. (ii) For product W, labour requirement = 2 hours, raw material = 2 kg, profit = ₹35. Profit per labour hour for W = 35/2 = 17.5, which is greater than X’s 16. Also, raw material remains redundant because all products have raw material per labour hour ≤ 1.2, so raw use ≤ 720 kg < 800 kg. Thus produce W using all labour: 2w ≤ 600 ⇒ w = 300. Raw material used = 2(300) = 600 ≤ 800. Profit = 300 × 35 = ₹10,500. This exceeds the old profit ₹9,600 by ₹900 per week. Alternatively, current labour shadow price = ₹16/hour and raw-material shadow price = ₹0/kg, so opportunity cost of one W = 2(16) + 2(0) = ₹32 < ₹35. Yes, include W. New optimal mix: 300 units of W, profit = ₹10,500 per week.

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